Question

Какова масса сульфата бария выпадает в осадок, если в реакцию вступает 9,8 г. cерной кислоты и 50 г. хлорида бария.

Answer 1

BaCl2+H2SO4=BaSO4+2HCl
M(BaCl2)=137+35.5*2=208
M(H2SO4)=1*2+32+16*4=98
M(BaSO4)=037+32+16*4=233
находим изыток и недостаток. Расчет массы соли по недостатку
1)50/208=0,24
2) 9,8/98=0,1
при отношении реагентов 1:1, тогда на 0,1 кислоты необходимо 0,1 хлорида , а его 0,24 соль в избытке. расчет по кислоте 9,8*233/98=23,3

Answer 2

Calculation of Barium Sulfate Precipitate Mass

To calculate the mass of barium sulfate precipitate formed when 9.8 grams of sulfuric acid (H2SO4) reacts with 50 grams of barium chloride (BaCl2), we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

The balanced chemical equation for the reaction between sulfuric acid and barium chloride is as follows:

H2SO4 + BaCl2 -> BaSO4 + 2HCl

From the equation, we can see that the stoichiometric ratio between sulfuric acid and barium sulfate is 1:1. This means that for every 1 mole of sulfuric acid, 1 mole of barium sulfate is formed.

To determine the limiting reactant, we need to compare the number of moles of each reactant. Let's calculate the number of moles for sulfuric acid and barium chloride:

Number of moles of sulfuric acid (H2SO4): Molar mass of H2SO4 = 1(2) + 32 + 16(4) = 98 g/mol Number of moles = mass / molar mass = 9.8 g / 98 g/mol = 0.1 mol

Number of moles of barium chloride (BaCl2): Molar mass of BaCl2 = 137 + 35.5(2) = 208 g/mol Number of moles = mass / molar mass = 50 g / 208 g/mol = 0.24 mol

Since the stoichiometric ratio between sulfuric acid and barium sulfate is 1:1, we can see that 0.1 moles of sulfuric acid will react with 0.1 moles of barium chloride to form 0.1 moles of barium sulfate.

The molar mass of barium sulfate (BaSO4) is 233.4 g/mol. Therefore, the mass of barium sulfate precipitate formed can be calculated as follows:

Mass of barium sulfate = number of moles * molar mass Mass of barium sulfate = 0.1 mol * 233.4 g/mol = 23.34 g

Therefore, the mass of barium sulfate precipitate formed when 9.8 grams of sulfuric acid and 50 grams of barium chloride react is approximately 23.34 grams.